The equilateral triangle ABP lies inside the square ABCD. Find `angle CPD`.

Options

A. 150°
B. 145°
C. 155°
D. 160°

Answer

To find the angle \( \angle CPD \) in the given configuration, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**: We have a square \( ABCD \) and an equilateral triangle \( ABP \) inside it. The vertices of the square are \( A, B, C, D \) in clockwise order.

2. **Identify Angles in the Equilateral Triangle**: Since \( ABP \) is an equilateral triangle, all its angles are \( 60^\circ \). Therefore, \( \angle APB = 60^\circ \).

3. **Determine Angles in the Square**: The angles in square \( ABCD \) are all \( 90^\circ \). Specifically, \( \angle ABC = 90^\circ \).

4. **Calculate \( \angle BPC \)**: Since \( \angle APB = 60^\circ \) and \( \angle ABC = 90^\circ \), we can find \( \angle BPC \) using the angle sum property:
\[ \angle BPC + \angle APB + \angle ABC = 180^\circ \]
\[ \angle BPC + 60^\circ + 90^\circ = 180^\circ \]
\[ \angle BPC = 180^\circ - 150^\circ = 30^\circ \]

5. **Use Isosceles Triangle Properties**: Since \( PB = AB \) (sides of the equilateral triangle), we have \( PB = BC \). Thus, \( \triangle BPC \) is isosceles with \( PB = BC \). Therefore, \( \angle BPC = \angle BCP = 30^\circ \).

6. **Calculate Remaining Angles**: Now, in triangle \( BPC \):
\[ \angle BPC + \angle BCP + \angle CPB = 180^\circ \]
\[ 30^\circ + 30^\circ + \angle CPB = 180^\circ \]
\[ \angle CPB = 180^\circ - 60^\circ = 120^\circ \]

7. **Determine \( \angle APD \)**: Since \( APD \) is also an isosceles triangle (with \( AP = AD \)), we have:
\[ \angle APD = \angle ADP = 75^\circ \]

8. **Find \( \angle CPD \)**: The sum of angles around point \( P \) is \( 360^\circ \):
\[ \angle CPD + \angle APD + \angle BPC + \angle APB = 360^\circ \]
\[ \angle CPD + 75^\circ + 30^\circ + 60^\circ = 360^\circ \]
\[ \angle CPD + 165^\circ = 360^\circ \]
\[ \angle CPD = 360^\circ - 165^\circ = 195^\circ \]

### Final Answer:
\[ \angle CPD = 195^\circ \]